Question
When you move up from the surface of the earth, the gravitation is reduced as $g=9.807-3.32 \times$ $10^{-6} z,$ with $z$ being the elevation in meters. By what percentage is the weight of an airplane reduced when it cruises at $11000 \mathrm{~m} ?$
Step 1
We can do this by substituting the value of $z$ into the given equation for $g$: \[g = 9.807 - 3.32 \times 10^{-6} \times z\] Substituting $z = 11000$ m, we get: \[g = 9.807 - 3.32 \times 10^{-6} \times 11000 = 9.7704 \, \text{m/s}^2\] Show more…
Show all steps
Your feedback will help us improve your experience
Mahnoor Khan and 99 other educators are ready to help you.
Ask a new question
Labs
Want to see this concept in action?
Explore this concept interactively to see how it behaves as you change inputs.
Key Concepts
Recommended Videos
When you move up from the surface of the earth, the gravitation is reduced as $g=9.807-3.32 \times$ $10^{-6} z,$ with $z$ as the elevation in meters. By how many percent is the weight of an airplane reduced when it cruises at $11000 \mathrm{m} ?$
If the variation of the acceleration of gravity, in $\mathrm{m} / \mathrm{s}^{2}$, with elevation $z$, in $\mathrm{m}$, above sea level is $g=9.81-\left(3.3 \times 10^{-6}\right) z$, determine the percent change in weight of an airliner landing from a cruising altitude of $10 \mathrm{~km}$ on a runway at sea level.
Getting Started: Introductory Concepts and Definitions
Problems: Developing Engineering Skills
The value of the gravitational acceleration $g$ decreases with elevation from $9.807 \mathrm{~m} / \mathrm{s}^{2}$ at sea level to $9.767 \mathrm{~m} / \mathrm{s}^{2}$ at an altitude of $13,000 \mathrm{~m}$, where large passenger planes cruise. Determine the percent reduction in the weight of an airplane cruising at $13,000 \mathrm{~m}$ relative to its weight at sea level.
Transcript
Watch the video solution with this free unlock.
EMAIL
PASSWORD